Let $Q$ be the midpoint of $AC$ and $N$ the intersection of $AD$ and $PQ$. Then $N$ is the midpoint of $AD$. As $DE$ is perpendicular to $AT$, being $E$ the intersection point of $DI$ and $AT$, and as $OA$ is perpendicular to $AT$, we get that $DE$ is parallel to $OA$, and so the angles $OAN$ and $ADE$ are equal. Indeed, $\angle OAQ = \angle BAH$ because $\angle OAQ = \frac{180^\circ - \angle COA}{2} = 90^\circ - \frac{\angle COA}{2} = 90^\circ - B = \angle BAH$. Moreover, $\angle BAD = \angle DAC$ and hence $\angle OAN = \angle HAD = \angle ADE$. As a consequence, triangles $ADE$ and $DAH$ are congruent.

In particular angle $DAT$ equals to angle $HAD$, that is, $ATD$ is isosceles and point $I$ is the orthocenter of $\triangle ABC$. So, $TI$ is perpendicular to $AD$, and the intersection point of $TI$ and $AD$ is the midpoint of $AD$, say $N$. The four points $M, N, I, T$ are collinear. We will apply the Ceva theorem in the triangle $APT$ with the cevians $PN, AD$ and $TM$. We get
$$\frac{FP}{FT} \cdot 1 \cdot \frac{MA}{PM} \Leftrightarrow \frac{PF}{TF} = \frac{MP}{MA}$$

(Observe that $NP$ cuts $AT$ in its midpoint) So, $MF$ is parallel to $AT$, and from this $MF$ is perpendicular to $AO$, as claimed.